To find the shoe size that puts a male in the lowest \(5\%\) of shoe sizes, we need to determine the \(5^{th}\) percentile of the normal distribution with mean \(\mu = 10.5\) and standard deviation \(\sigma = 1.36\).
Using the z-score for the \(5^{th}\) percentile, we have:
\[
z_{0.05} \approx -1.6449
\]
The corresponding shoe size \(X\) can be calculated using the formula:
\[
X = \mu + z \cdot \sigma
\]
Substituting the values:
\[
X = 10.5 + (-1.6449) \cdot 1.36 \approx 8.26
\]
Thus, the shoe size for the lowest \(5\%\) is:
\[
\boxed{8.26}
\]
Next, we need to find the probability that a male has a shoe size of \(14.5\) or larger. This can be expressed as:
\[
P(X \geq 14.5)
\]
First, we calculate the z-score for \(X = 14.5\):
\[
z = \frac{X - \mu}{\sigma} = \frac{14.5 - 10.5}{1.36} \approx 2.9412
\]
Next, we find the probability using the cumulative distribution function \(\Phi\):
\[
P(X \geq 14.5) = 1 - \Phi(z)
\]
Using the z-score calculated:
\[
P(X \geq 14.5) = 1 - \Phi(2.9412) \approx 1 - 0.9984 = 0.0016
\]
Thus, the probability that a male has a shoe size of \(14.5\) or larger is:
\[
\boxed{0.0016}
\]
- Shoe size for the lowest \(5\%\): \(\boxed{8.26}\)
- Probability of shoe size \(14.5\) or larger: \(\boxed{0.0016}\)